FINAL JEE–MAIN EXAMINATION – APRIL, 2024
(Held On Monday 08th April, 2024) TIME : 3 : 00 PM to 6 : 00 PM
NOON
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. If the image of the point (-4, 5) in the line
x + 2y = 2 lies on the circle (x + 4)2 + (y-3)2 = r2,
then r is equal lo :
(1) 1 (2) 2
(3) 75 (4) 3
Ans. (2)
Sol. Image of point (–4, 5)
1 1 1 1
22
x x y y ax by c
2
ab ab
Line : x + 2y – 2 = 0
22
x 4 y 5 4 10 2
2
12 12
8
5
8 28
x4
55
16 9
y5
55
Point lies on circle (x + 4)2 + (y – 3)2 = r2
2
2
64 9 3r
25 5
2
100 r , r 2
25
2. Let
ˆˆˆ
a i 2 j 3k
,
ˆˆˆ
b 2i 3j 5k
and
ˆˆ ˆ
c 3i j k
be three vectors. Let
r
be a unit
vector along
bc
. If
r . a 3
, then 3 is equal
to :
(1) 27 (2) 25
(3) 25 (4) 21
Ans. (2)
Sol.
r k b c
r . a 3
r.a k b.a c.a
3 = k(2 + 6 – 15 + 3 –2 + 3)
3 = k(–6 + 3) …(1)
ˆˆ ˆ
r k(5i 2j (5 )k)
2
r k 25 4 25 10 1
…(2)
31
k6 3 2
put in (2)
4 + 2 – 4 = 54 + 2 – 10
6 = 50
3 = 25
3. If a, b, c and
bc
a c 0
ab
, then
ab
a b c
is equal to :
(1) 2 (2) 3
(3) 0 (4) 1
Ans. (3)
Sol. R1 R 1 – R 2, R 2 R 2 – R 3
a b 0
0 b c 0
ab
(–a) ((–b) – b(c–)) – (b – ) (– a(c - )) = 0
( – a) ( – b) – b( – a) (c – ) + a(b - ) (c – )
ba
0
c b a
4. In an increasing geometric progression ol positive
terms, the sum of the second and sixth terms is
70
3
and the product of the third and fifth terms is
49. Then the sum of the 4th, 6th and 8th terms is :-
(1) 96 (2) 78
(3) 91 (4) 84
Ans. (3)
Sol.
26
70
TT 3
ar + ar5 =
70
3
T3 . T5 = 49
ar2 . ar4 = 49
a2r6 = 49
ar3 = +7,
3
7
ar
ar(1 + r4) =
70
3
42
2
7 70
(1 r ) , r t
3
r
2
1 10
(1 t )
t3
3t2 – 10t + 3 = 0
1
t 3, 3
Increasing G.P. r2 = 3,
r3
T4 + T6 + T8
= ar3 + ar5 + ar7
= ar3(1 + r2 + r4)
= 7(1 + 3 + 9) = 91
5. The number of ways five alphabets can be chosen
from the alphabets of the word MATHEMATICS,
where the chosen alphabets are not necessarily
distinct, is equal to :
(1) 175 (2) 181
(3) 177 (4) 179
Ans. (4)
Sol. AA, MM, TT, H, I, C, S, E
(1) All distinct
8C5 56
(2) 2 same, 3 different
3C × 7C3 105
(3) 2 same Ist kind, 2 same 2nd kind, 1 different
3C2 × 6C1 18
Total 179
6. The sum of all possible values of [–, 2], for
which
1 i cos
1 2i cos
is purely imaginary, is equal
to
(1) 2 (2) 3
(3) 5 (4) 4
Ans. (2)
Sol.
1 i cos
Z1 2i cos
ZZ
1 icos 1 icos
1 2i cos 1 2i cos
(1+ icos)
1 2icos
=–(1 – 2i cos)
1 icos
(1+icos) (1 + 2icos) = –(1 – 2icos) (1 – icos)
1 + 3icos – 2cos2 = –(1 – 3icos – 2cos2)
2 – 4cos2 = 0
cos2 =
1
2
3 3 5 7
, , , , ,
4 4 4 4 4 4
sum = 3
7. If the system of equations x + 4y – z = ,
7x + 9y + z = –3, 5x + y + 2z = –1 has infinitely
many solutions, then (2. + 3) is equal to :
(1) 2 (2) –3
(3) 3 (4) –2
Ans. (2)
Sol.
1 4 1
7 9 0
5 1 2
(18–) – 4(14–5) – (7 – 45) = 0 = 0
= x = y = z = 0 (For infinite solution)
x
41
3 9 0
1 1 2
(18 – ) – 4(–6 + ) –1(–3 + 9) = 0
18 + 24 – 6 = 0 = –1
8. If the shortest distance between the lines
x y 4 z 3
2 3 4
and
x 2 y 4 z 7
4 6 8
is
13
29
, then a value
of is :
(1) –
13
25
(2)
13
25
(3) 1 (4) –1
Ans. (3)
Sol.
1
1
22
ˆˆ
b 2i 3j 4k
ˆ ˆ ˆ ˆ
ˆˆ
r i 4 j 3k (2i 3 j 4k) ˆˆˆ
a i 4 j 3k
ˆ ˆ ˆ ˆ
ˆˆ
r 2i 4 j 7k (2i 3j 4k) ˆˆˆ
a 2i 4 j 7k
21
b (a a ) 13
Shortest dist. b29
ˆ ˆ ˆ
ˆˆ
2i 3j 4k (2 )i 4k 13
29 29
ˆ ˆ ˆ
ˆ
8j 3(2 )k 12i 4(2 ) j 13
ˆˆ ˆ
12i 4 j (3 6)k 13
144 + 16 2 + (3 – 6)2 = 169
162 + (3 – 6)2 = 25 = = 1
9. If the value of
3cos36 5sin18
5cos36 3sin18
is
a 5 b
c
,
where a, b, c are natural numbers and gcd(a, c) = 1,
then a + b + c is equal to :
(1) 50 (2) 40
(3) 52 (4) 54
Ans. (3)
Sol.
3 5 1 5 1
58 5 2
44
2 5 8
5 1 5 1
53
44
4 5 1 5 4
5 4 5 4
20 16 5 5 4
11
17 5 24
11
a = 17, b = 27, c = 11
a + b + c = 52
10. Let y = y(x) be the solution curve of the
differential equation secy
dy
dx
+ 2xsiny = x3cosy,
y(1) = 0. Then
y3
is equal to :
(1)
3
(2)
6
(3)
4
(4)
12
Ans. (3)
Sol. sec2y
dy
dx
+ 2xsiny secy = x3cosy secy
sec2y
dy
dx
+ 2xtany = x3
tany = t
2dy dt
sec y dx dx
3
dt 2xt x
dx
, If
2
2 x dx x
ee
22
x 3 x
te x .e dx c
x2 = Z
Z Z Z Z
11
t.e e .ZdZ e .Z e c
22
2
2x
2tan y (x 1) 2ce
y(1) = 0 c = 0
y( 3) 4
11. The area of the region in the first quadrant inside
the circle x2 + y2 = 8 and outside the pnrabola
y2 = 2x is equal to :
(1)
1
23
(2)
2
3
(3)
2
23
(4)
1
3
Ans. (2)
Sol.
(2 ,0)
x=2
Required area = Ar(circle from 0 to 2) –
ar(para from 0 to 2)
22
2
0
0
8 x dx 2x dx
2
2
21
0
0
x 8 x x x
8 x sin 2
2 2 3 / 2
22
1
2 8 2 2 2
8 4 sin 2 2 0
2 2 3
22
82
2 4 . 4 3 3
12. If the line segment joining the points (5, 2) and
(2, a) subtends an angle
4
at the origin, then the
absolute value of the product of all possible values
of a is :
(1) 6 (2) 8
(3) 2 (4) 4
Ans. (4)
Sol.
B(2, a)
/4
A(5,2)
O
OA
2
m5
OB
a
m2
2a
tan 52
4
4 5a
110 2a
4 – 5a = ±(10 + 2a)
4 – 5a = 10 + 2a 4 – 5a = –10 – 2a
7a + 6 = 0 3a = 14
6
a7
14
a3
6 14 4
73
13. Let
ˆˆˆ
a 4i j k
,
ˆˆˆ
b 11i j k
and
c
be
a vector such that
a b c c 2a 3b
.
If
2a 3b .c 1670
, then
2
| c |
is equal to :
(1) 1627 (2) 1618
(3) 1600 (4) 1609
Ans. (2)
Sol.
a b c c 2a 3b 0
a b c 2a 3b c 0
a b 2a 3b) c 0
c (4b a)
ˆ ˆ ˆ ˆ
ˆˆ
44i 4 j 4k 4i j k
ˆˆ ˆ
40i 3j 3k
Now
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ
8i 2 j 2k 33i 3j 3k . (40i 3j 3k) 1670
ˆ ˆ ˆ ˆ
ˆˆ
41i 5 j 5k .(40i 3 j 3k) 1670
)
(1640 + 15 + 15) = 1670 = 1
so
ˆˆˆ
c 40i 3j 3k
2
c 1600 9 9 1618
14. If the function f(x) = 2x3 – 9ax2 + 12a2x + 1, a > 0
has a local maximum at x = and a local
minimum x = 2, then and 2 are the roots of the
equation :
(1) x2 – 6x + 8 = 0 (2) 8x2 + 6x – 8 = 0
(3) 8x2 – 6x + 1 = 0 (4) x2 + 6x + 8 = 0
Ans. (1)
Sol.
f'(x) = 6x2 – 18ax + 12a2 = 0
2
+ 2 = 3a & × 2 = 2a2
( + 2)3 = 27a3
2a2 + 4a4 + 3(3a) (2a2) = 27a3
2 + 4a2 + 18a = 27a
4a2 – 9a + 2 = 0
4a2 – 8a – a + 2 = 0
(4a – 1) (a – 2) = 0 a = 2
so 6x2 – 36x + 48 = 0
x2 – 6x + 8 = 0 (1)
If we take
1
a4
then
1
2
which is not possible
15. There are three bags X, Y and Z. Bag X contains 5
one-rupee coins and 4 five-rupee coins; Bag Y
contains 4 one-rupee coins and 5 five-rupee coins
and Bag Z contains 3 one-rupee coins and
6 five-rupee coins. A bag is selected at random and
a coin drawn from it at random is found to be a
one-rupee coin. Then the probability, that it came
from bag Y, is :
(1)
1
3
(2)
1
2
(3)
1
4
(4)
5
12
Ans. (1)
Sol. X Y Z
5 one & 4 five 4 one & 5 five 3 one & 6 five
4 / 9 4 1
P5 / 9 4 / 9 3 / 9 12 3
16. Let
log 4
x
edx
6
e1
. Then e and e– are the
roots of the equation :
(1) 2x2 – 5x + 2 = 0 (2) x2 – 2x - 8 = 0
(3) 2x2 – 5x – 2 = 0 (4) x2 + 2x – 8 = 0
Ans. (1)
Sol.
log 4
x
edx
6
e1
Let ex – 1 = t2
ex dx = 2t dt
2
2dt
t1
= 2 tan–1t
4
e
log
1x
2 tan e 1
11
2 tan 3 tan e 1 6
1
tan e 1
3 12
1
tan e 1 4
e2
1
e2
21
x 2 x 1 0
2
2x2 – 5x + 2 = 0
17. Let
a if a x 0
(x) x a if 0 x a
f
where a > 0 and g(x) = (f |x| ) – | f (x)| )/2.
Then the function g : [ –a, a] [ –a, a] is
(1) neither one-one nor onto.
(2) both one-one and onto.
(3) one-one.
(4) onto
Ans. (1)
Sol. y = f(x)
–a
O
a
a
2a
–a
y = f|x|
–a
a
a
2a
y = |f(x)|
|
–a
a
2a
a
g(x) =
f(| x |) | f(x) |
2
–a
O
a/2
a
18. Let A= {2, 3, 6, 8, 9, 11} and B = {1, 4, 5, 10, 15}
Let R be a relation on A × B define by (a, b)R(c, d)
if and only if 3ad – 7bc is an even integer. Then
the relation R is
(1) reflexive but not symmetric.
(2) transitive but not symmetric.
(3) reflexive and symmetric but not transitive.
(4) an equivalence relation.
Ans. (3)
Sol. A = {2, 3, 6, 8, 9, 11} (a, b)R (c, d)
B = {1, 4, 5, 10, 15} 3ad – 7bc
Reflexive : (a, b) R(a, b)
3ab – 7ba = – 4ab always even so it is reflexive.
Symmetric : If 3ad – 7bc = Even
Case-I : odd odd
Case-II : even even
(c, d) R(a, b) 3bc – 3ab
Case-I : odd odd
Case-II : even even
so symmetric relation
Transitive :
Set (3, 4)R (6, 4) Satisfy relation
Set (6, 4)R(3, 1) Satisfy relation
but (3, 4) R(3, 1) does not satisfy relation
so not transitive.
19. For a, b > 0, let
22
tan((a 1)x) b tan x , x 0
x
3 , x 0
(x)
ax b x ax , x 0
b a x x
f
be a continous function at x = 0. Then
b
a
is equal
to
(1) 5 (2) 4
(3) 8 (4) 6
Ans. (4)
Sol.
x0
lim f(x) f(0) 3
22
x0
ax b x ax
lim 3
b a x x
22
3/2 2 2
x0
ax b x ax
lim
b a x ax b x ax
2
2
x0
b
lim
b a a b x a
bb
3
2a
a.2 a
b6
a
20. If the term independent of x in the expansion of
10
2
3
1
ax 2x
is 105, then a2 is equal to :
(1) 4 (2) 9
(3) 6 (4) 2
Ans. (1)
Sol.
10
2
3
1
ax 2x
General term =
r
10 r
10 2
r3
C ax 2x
20 – 2r – 3r = 0
r = 4
10 3
4
1
C a . 105
16
a3 = 8
a2 = 4
SECTION-B
21. Let A be the region enclosed by the parabola
y2 = 2x and the line x = 24. Then the maximum
area of the rectangle inscribed in the region A is
_________ .
Ans. (128)
Sol.
x= 24
(24, b)
A =
2
b
2 24 .b
2
dA 0
db
b = 4
A = 2(24 – 8)4
= 128
22. If
tan x x
x0
ee
lim tan x x
and
cot x
x0
1
2
lim(1 sin x)
are the roots of the
quadratic equation ax2 + bx –
e
= 0, then 12
loge(a + b) is equal to _________ .
Ans. (6)
Sol.
tan x x
x
x0
e1
lim e tan x x
= 1
1cot x
2
x0
lim (1 sin x)
= e1/2
2
x 1 e e 0
ax2 + bx –
e
= 0
On comparing
a = –1, b =
e
+1
12 n(a + b) = 12 ×
1
2
= 6
23. Let S be the focus of the hyperbola
22
xy
1
35
,
on the positive x-axis. Let C be the circle with its
centre at
A 6, 5
and passing through the
point S. if O is the origin and SAB is a diameter of
C then the square of the area of the triangle OSB is
equal to -
Ans. (40)
Sol.
B
A
O
Area =
1
2
(OS) h =
1
2
8
25
=
40
24. Let P(,,) be the image of the point Q(l, 6, 4) in
the line
x y 1 z 2
1 2 3
. Then 2 + + is
equal to _______ .
Ans. (11)
Sol.
Q(1, 6, 4)
P, , )
A(t, 2t + 1, 3t + 2)
ˆˆˆ
QA (t 1)i (2t 5)j (3t 2)k
.
QA b 0
(t – 1) + 2(2t – 5) + 3(3t – 2) = 0
14t = 17
20
14
12
14
102
14
2 =
154
14
= 11
25. An arithmetic progression is written in the
following way
2
5 8
11 14 17
20 23 26 29
_______________________________________
_______________________________________
The sum of all the terms of the 10th row is ______ .
Ans. (1505)
Sol. 2, 5, 11, 20, …..
General term =
2
3n 3n 4
2
10
3(100) 3(10) 4
T2
= 137
10 terms with c.d. = 3
10
sum 2(137) 9(3)
2
= 1505
26. The number of distinct real roots of the equation
|x + 1| |x + 3| – 4 |x + 2| + 5 = 0, is ________ .
Ans. (2)
Sol. |x + 1| |x + 3| - 4|x + 2| + 5 = 0
case–1
x –3
(x + 1) (x + 3) + 4(x + 2) + 5 = 0
x2 + 4x + 3 + 4x + 8 + 5 = 0
x2 + 8x + 16 = 0
(x + 4)2 = 0
x = –4
case-2
–3 x –2
–x2 – 4x – 3 + 4x + 8 + 5 = 0
–x2 + 10 = 0
x 10
case-3
–2 x –1
–x2 – 4x – 3 – 4x – 8 + 5 = 0
–x2 – 8x – 6 = 0
x2 + 8x + 6 = 0
8 2 10
x 4 10
2
case-4
x–1
x2 + 4x + 3 – 4x – 8 + 5 = 0
x2 = 0
x = 0
No. of solution = 2
27. Let a ray of light passing through the point (3, 10)
reflects on the line 2x + y = 6 and the reflected ray
passes through the point (7, 2). If the equation of
the incident ray is ax + by + l = 0, then
a2 + b2 + 3ab is equal to_.
Ans. (1)
Sol.
A(3,10)
ax+by+1=c
B(7,2)
2x+y–6=0
B'
For B'
x 7 y 2 14 2 6
2
2 1 5
x 7 y 2 4
21
x = –1 y = –2 B'(–1, –2)
incident ray AB'
MAB' = 3
y + 2 = 3(x + 1)
3x – y + 1 = 0
a = 3 b = –1
a2 + b2 + 3ab = 9 + 1 – 9 = 1
28. Let a, b, c N and a < b < c. Let the mean, the
mean deviation about the mean and the variance of
the 5 observations 9, 25, a, b, c be 18, 4 and
136
5
,
respectively. Then 2a + b – c is equal to _______ .
Ans. (33)
Sol. a, b, c N a < b < c
9 25 a b c
x mean 18
5
a + b + c = 56
Mean deviation =
i
| x x | 4
n
= 9 + 7 + |18 – a| + |18 – b| + |18 – c| = 20
= |18 – a| + |18 – b| + |18 – c| = 4
Variance =
2
i
| x x | 136
n5
= 81 + 49 + |18 – a|2 + |18 – b|2 + |18 – c|2 = 136
= (18 – a)2 + (18 – b)2 + (18 – c)2 = 6
Possible values (18–a)2 = 1, (18–b)2=1 (18–c)2=4
a < b < c
so 18–a=1 18–b=–1 18–c=–2
a=17 b=19 c=20
a + b + c = 56
2a + b – c 34 = 19 – 20 = 33
29. Lei |x| = |y|exy–, , N be the solution of the
differential equation xdy – ydx + xy(xdy+ydx) = 0,
y(l) = 2. Then + is equal to _
Ans. (4)
Sol. a|x| = |y| eyx-, a, b N
xdy – ydx + xy(xdy + ydx) = 0
dy dx (xdy ydx) 0
yx
n|y| – n|x| + xy = c
y(1) = 2
n|2| – 0 + 2 = c
c = 2 + n2
n|y| – n|x| + xy = 2 + n2
n|x| =
y
n2
– 2 + xy
xy 2
y
xe
2
2|x| = |y|exy-2
= 2 = 2 + = 4
30. If
B
46
5
1 x 1
dx A C
x3
(x 1) (x 3)
,
where C is the constant of integration, then the
value of + + 20AB is _______ .
Ans. (7)
Sol.
B
46
5
1 x 1
dx A C
x3
(x 1) (x 3)
4/5 6/5
1
I dx
(x 1) (x 3)
4/5
2
1
I dx
x1 (x 3)
x3
x1
x3
= t
2
4dx dt
(x 3)
t–4/5+1
I =
1/5
4/5
1 1 1 t
dt c
4 4 1 / 5
t
I =
1/5
5 x 1 C
4 x 3
A =
5
4
= = 1 B =
1
5
+ + 20AB = 2 + 20 ×
51 7
45
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31.
µ=0.5
30º
µ=0
k = 100 N/m
2m
A block is simply released from the top of an
inclined plane as shown in the figure above. The
maximum compression in the spring when the
block hits the spring is :
(1)
6m
(2) 2 m
(3) 1 m (4)
5m
Ans. (2)
Sol. wg + wFr + ws = KE
5 × 10 × 5 – 0.5 × 5 × 10 × x –
1
2
Kx2 = 0 – 0
250 = 25 x + 50 x2
2x2 + x – 10 = 0
x = 2
32. In a hypothetical fission reaction
92X236 56Y141 +36Z92 + 3R
The identity of emitted particles (R) is :
(1) Proton (2) Electron
(3) Neutron (4) -radiations
Ans. (3)
Sol. Z in LHS = 92
Z in RHS = 56 + 36 = 92
A in LHS = 236
A in RHS = 141 + 92 = 233
So 3 neutrons are released.
33. If is the permittivity of free space and E is the
electric field, then
2
0E
has the dimensions :
(1) [Mo L–2 T A] (2) [M L–1 T–2]
(3) [M–1 L–3 T4 A2] (4) [M L2 T–2]
Ans. (2)
Sol. E =
2
KQ
R
E =
2
o
Q
4R
o =
2
Q
4 R E
Now, oE2 =
2
Q
4 R E
.E2 =
2
Q
4R
.E
[oE2] =
2
QE
R
=
2
[Q][E]
[R ]
=
2
[Q] [W]
[Q][R]
[R ]
=
3
[W]
[R ]
=
22
3
ML T
L
=
12
ML T
34. The position of the image formed by the
combination of lenses is :
f1=10 cm
f2=–10
f3=30 cm
30cm
5cm
10cm
(1) 30 cm (right of third lens)
(2) 15 cm (left of second lens)
(3) 30 cm (left of third lens)
(4) 15 cm (right of second lens)
Ans. (1)
Sol. For lens 1 : f1 = 10, u = –30, v = ?
v =
uf
uf
=
–30 10
–30 10
= 15
For lens 2 : f1 = –10, u = 10, v = ?
v =
uf
uf
=
10 10
10 10
=
For lens 3 : f = 30, u = –, v = ?
So v will be 30.
35. A plane progressive wave is given by
y = 2 cos 2(330 t – x) m. The frequency of the
wave is :
(1) 165 Hz (2) 330 Hz
(3) 660 Hz (4) 340 Hz
Ans. (2)
Sol. y = 2 cos 2(330 t – x) m
y = Acos (t – kx)
by comparing = 2× 330
2f = 2 × 330
f = 330
36. A thin circular disc of mass M and radius R is
rotating in a horizontal plane about an axis passing
through its centre and perpendicular to its plane
with angular velocity . If another disc of same
dimensions but of mass
M2
is placed gently on
the first disc co-axially, then the new angular
velocity of the system is :
(1)
4
5
(2)
5
4
(3)
2
3
(4)
3
2
Ans. (3)
Sol. I1 = I2
22
2
MR 3 MR
2 2 2
=
2
3
37. A cube of ice floats partly in water and partly in
kerosene oil. The radio of volume of ice immersed
in water to that in kerosene oil (specific gravity of
Kerosene oil = 0.8, specific gravity of ice = 0.9)
Kerosene
oil
Water
(1) 8 : 9 (2) 5 : 4
(3) 9 : 10 (4) 1 : 1
Ans. (4)
Sol. v1 = volume immersed in water.
v2 = volume immersed in oil.
v1 w g + v2 o g = (v1 + v2) c g
v1 +
2o
w
v
(v1 + v2)
c
w
= v1 + 0.8 v2 = 0.9 v1 + 0.9 v2
= 0.1 v1 = 0.1 v2
v1 : v2 = 1 : 1
38. Given below are two statements :
Statement (I) : The mean free path of gas
molecules is inversely proportional to square of
molecular diameter.
Statement (II) : Average kinetic energy of gas
molecules is directly proportional to absolute
temperature of gas.
In the light of the above statements, choose the
correct answer from the option given below:
(1) Statement I is false but Statement II is true.
(2) Statement I is true but Statement II is false.
(3) Both Statement I and Statement II are false
(4) Both Statement I and Statement II are true.
Ans. (4)
Sol.
2
A
RT
2 d N P
KE =
fnRT
2
39. Two satellite A and B go round a planet in circular
orbits having radii 4 R and R respectively. If the
speed of A is 3v, the speed of B will be :
(1)
4
3v
(2) 3v
(3) 6v (4) 12v
Ans. (3)
Sol. v =
GM
R
AB
BA
vR R1
v R 4R 2
BA
v 2v 6v
40. A long straight wire of radius a carries a steady
current I. The current is uniformly distributed
across its cross section. The ratio of the magnetic
field at
a
2
and 2a from axis of the wire is :
(1) 1 : 4 (2) 4 : 1
(3) 1 : 1 (4) 3 : 4
Ans. (3)
Sol.
1o
aI
B 2 µ
24
o
1
µI
B4a
2o
B 2 2a µ I
o
2
µI
B4a
41. The angle of projection for a projectile to have
same horizontal range and maximum height is :
(1) tan–1 (2) (2) tan–1 (4)
(3)
11
tan 4
(4)
11
tan 2
Ans. (2)
Sol.
2 2 2
u sin2 u sin
g 2g
4sin cos = sin2
4 = tan
42. Water boils in an electric kettle in 20 minutes after
being switched on. Using the same main supply,
the length of the heating element should be
………….. to …………. times of its initial length
if the water is to be boiled in 15 minutes.
(1) increased,
3
4
(2) increased,
4
3
(3) decreased,
3
4
(4) decreased,
4
3
Ans. (3)
Sol.
2
v
PR
,
RA
1
P
1 2 2
2 1 1
Pt 15
P t 20
2 =
3
4
1
43. A capacitor has air as dielectric medium and two
conducting plates of area 12 cm2 and they are
0.6 cm apart. When a slab of dielectric having area
12 cm2 and 0.6 cm thickness is inserted between
the plates, one of the conducting plates has to be
moved by 0.2 cm to keep the capacitance same as
in previous case. The dielectric constant of the slab
is : (Given = 8.834 × 10–12 F/m)
(1) 1.50 (2) 1.33
(3) 0.66 (4) 1
Ans. (1)
Sol.
oo
AA
d
d0.2 k
0.6 = 0.2 +
0.6
k
k =
3
2
44. A given object takes n times the time to slide down
45° rough inclined plane as it takes the time to
slide down an identical perfectly smooth 45°
inclined plane. The coefficient of kinetic friction
between the object and the surface of inclined
plane is :
(1)
2
1
1n
(2) 1 – n2
(3)
2
1
1n
(4)
2
1n
Ans. (1)
Sol.
= 45º
Case-1 : No friction
a = g sin
=
2
1
1(gsin )t
2
t1 =
2
gsin
Case-2 : With friction
a = g sin – µg cos
=
2
2
1(gsin µgcos )t
2
22
n
gsin µgcos gsin
µ = 1 –
2
1
n
45. A coil of negligible resistance is connected in
series with 90 resistor across 120 V, 60 Hz
supply. A voltmeter reads 36 V across resistance.
Inductance of the coil is :
(1) 0.76 H (2) 2.86 H
(3) 0.286 H (4) 0.91 H
Ans. (1)
Sol.
36V
R
L
120V/60Hz
36 = Irms R
36 =
22
L
120 R
XR
R = 90 36 =
22
L
120 90
X 90
22
L
X 90
= 300
2
L
X
= 81900
L
X
= 286.18
L = 286.18
L =
286.18
376.8
L = 0.76 H
46. There are 100 divisions on the circular scale of a
screw gauge of pitch 1 mm. With no measuring
quantity in between the jaws, the zero of the
circular scale lies 5 divisions below the reference
line. The diameter of a wire is then measured using
this screw gauge. It is found the 4 linear scale
divisions are clearly visible while 60 divisions on
circular scale coincide with the reference line. The
diameter of the wire is :
(1) 4.65 mm (2) 4.55 mm
(3) 4.60 mm (4) 3.35 mm
Ans. (2)
Sol. Least count =
1mm 0.01mm
100
zero error = +0.05 mm
Reading = 4 × 1 mm + 60 × 0.01 mm – 0.05 mm
= 4.55 mm
47. A proton and an electron have the same de Broglie
wavelength. If Kp and Ke be the kinetic energies of
proton and electron respectively. Then choose the
correct relation :
(1) Kp > Ke (2) Kp = Ke
(3) Kp = Ke2 (4) Kp < Ke
Ans. (4)
Sol. De Broglie wavelength of proton & electron =
=
h
p
pproton = pelectron
KE =
2
p
2m
KEproton < KEelectron
[Kp < Ke]
48. Least count of a vernier caliper is
20N
cm. The
value of one division on the main scale is 1 mm.
Then the number of divisions of main scale that
coincide with N divisions of vernier scale is :
(1)
2N 1
20N
(2)
2N 1
2
(3) (2N – 1) (4)
2N 1
2N
Ans. (2)
Sol. Least count of vernier calipers =
1cm
20N
Least count = 1 MSD – 1 VSD
let x no. of divisions of main scale coincides with
N division of vernier scale, then
1 VSD =
x 1mm
N
1cm
20N
= 1 mm –
x 1mm
N
1x
mm 1mm – mm
2N N
x =
1
12N
N
x =
2N 1
2
49. If Mo is the mass of isotope
12
5B
, Mp and Mn are
the masses of proton and neutron, then nuclear
binding energy of isotope is :
(1) (5 Mp + 7Mn – Mo)C2
(2) (Mo – 5Mp)C2
(3) (Mo – 12Mn)C2
(4) (Mo – 5Mp – 7Mn)C2
Ans. (1)
Sol. B.E. = mC2
(5 Mp + 7Mn – Mo)C2
50. A diatomic gas ( = 1.4) does 100 J of work in an
isobaric expansion. The heat given to the gas is :
(1) 350 J (2) 490 J
(3) 150 J (4) 250 J
Ans. (1)
For Isobaric process
w = Pv = nRT = 100 J
Q = u + w
Q =
FnR T nR T
2
f1 nR T
2
51 100
2
= 350 J
SECTION-B
51. The coercivity of a magnet is 5 × 103 A/m. The
amount of current required to be passed in a
solenoid of length 30 cm and the number of turns
150, so that the magnet gets demagnetised when
inside the solenoid is …………A.
Ans. (10)
Sol.
He
o
c
o
µ ni
Hµ
5 × 103 =
150 100 i
30
50
5
= i
I = 10
52. Small water droplets of radius 0.01 mm are formed
in the upper atmosphere and falling with a terminal
velocity of 10 cm/s. Due to condensation, if 8 such
droplets are coalesced and formed a larger drop,
the new terminal velocity will be ………..cm/s.
Ans. (40)
Sol. m = mass of small drop
M = mass of bigger drop
2
t
2 R ( )g
V9
8 m = M
8r3 = R3 R = 2R
as Vt × R2 Radius double so Vt becomes 4 time
4 × 10 = 40 cm/s
53. If the net electric field at point P along Y axis is
zero, then the ratio of
2
3
q
q
is
8
5x
,
where x = …………..
4cm
P
+q2
2cm
3cm
–q3
Ans. (5)
Sol.
4cm
+q2
2cm
3cm
–q3
Q2
Q1
25
P
20
3
2Kq
Kq cos cos
20 25
3
2Kq
Kq 44
20 25
20 25
2
3
q20 20 8
q 25 25 5x
8 25 25
x5 20 20
x = 5
54. A heater is designed to operate with a power of
1000 W in a 100 V line. It is connected in
combination with a resistance of 10 and a
resistance R, to a 100 V mains as shown in figure.
For the heater to operate at 62.5 W, the value of R
should be ……………. .
heater
B
C
R
100 V
10
Ans. (5)
Sol. Rheater =
22
V (100) 10
P 1000
For heater P =
2
VV PR
R
V =
62.5 10
V = 25 v
heater
R
100 V
10
10
iH
25V
iR
i1
75V
i1 =
75 7.5A,
10
iH =
25
10
= 2.5 A.
iR = i1 – iH = 5
V = IR
R =
25 5
5
55. An alternating emf E = 110
2
sin 100t volt is
applied to a capacitor of 2µF, the rms value of
current in the circuit is ………. mA.
Ans. (22)
Sol. C = 2µf ; E =
110 2
sin (100 t)
XC =
6
11
c100 2 10
=
10000 5000
2
io =
110 2
5000
irms =
110 2
5000 2
=
110 mA
5
= 22 mA
56. Two slits are 1 mm apart and the screen is located
1 m away from the slits. A light wavelength
500 nm is used. The width of each slit to obtain 10
maxima of the double slit pattern within the central
maximum of the single slit pattern is
……….. × 10–4 m.
Ans. (2)
Sol. d = 1 mm, D = 1m, = 500 nm
D 2 D
10 da
a =
d
5
=
–4
10 10 m
5
= 2 × 10–4
57. An object of mass 0.2 kg executes simple
harmonic motion along x axis with frequency of
25
Hz. At the position x = 0.04 m the object
has kinetic energy 0.5 J and potential energy 0.4 J.
The amplitude of oscillation is …………. cm.
Ans. (6)
Sol. Total energy = K.E. + P.E.
at x = 0.04 m, T.E. = 0.5 + 0.4 = 0.9 J
T.E = 1 m2A2 = 0.9
=
2
2
1 25
0.2 2 A 0.9
2
A = 0.06 m
A = 6 cm
58. A potential divider circuit is connected with a dc
source of 20 V, a light emitting diode of glow in
voltage 1.8 V and a zener diode of breakdown
voltage of 3.2 V. The length (PR) of the resistive
wire is 20 cm. The minimum length of PQ to just
glow the LED is …………. cm.
R
P
Q
20V
Ans. (5)
Sol.
R
P
Q
20V
1.8
3.2
PR = 20 cm
VPQ =
1
4
× RPR
min (PQ) =
120
4
= 5 cm
59. A body of mass M thrown horizontally with
velocity v from the top of the tower of height H
touches the ground at a distance of 100m from the
foot of the tower. A body of mass 2M thrown at a
velocity
2
v
from the top of the tower of height 4H
will touch the ground at a distance of ……….m.
Ans. (100)
Sol.
100m
x
M
v
H
4H
v
2
2M
2H 2(4H) 2H
100 ; x
g 2 g g
v
vv
x = 100
60. A circular table is rotating with an angular velocity
of rad/s about its axis (see figure). There is a
smooth groove along a radial direction on the
table. A steel ball is gently placed at a distance of
1m on the groove. All the surface are smooth. If
the radius of the table is 3 m, the radial velocity of
the ball w.r.t. the table at the time ball leaves the
table is
x2
m/s, where the value of x
is………..
1m
3m
Ans. (2)
Sol. ac = 2x
2
vdv x
dx
V3
2
01
vdv xdx
22
2
vx
22
22
22
v31
22
v =
22
x = 2
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. In qualitative test for identification of presence of
phosphorous, the compound is heated with an
oxidising agent. Which is further treated with nitric
acid and ammonium molybdate respectively. The
yellow coloured precipitate obtained is :
(1)
3 4 3
Na PO .12MoO
(2)
4 4 4 4
32
NH PO .12 NH MoO
(3)
4 4 3
3
NH PO .12MoO
(4)
4 4 3
MoPO .21NH NO
Ans. (3)
Sol.
PO43–
Or HPO4–
+ (NH4)2MoO4
Ammonium
Molybdate
(NH4)3PO4.12MoO3
Canary yellow ppt.
(Ammonium phopho
molybdate)
H+
62. For a reaction
12
KK
A B C
If the rate of formation of B is set to be zero then
the concentration of B is given by :
(1) K1K2[A] (2) (K1 – K2)[A]
(3) (K1 + K2)[A] (4) (K1/K2)[A]
Ans. (4)
Sol. Rate of formation of B is
12
dB k A k B
dt
0 = k1[A] – k2[B]
1
2
kAB
k
63. When and B are the wave functions of atomic
orbitals, then * is represented by :
(1) A – 2B (2) A – B
(3) A + 2B (4) A + B
Ans. (2)
Sol. Antibonding molecular orbitals are formed by
destructive interference of wave functions.
(ABMO) * = A – B
64. Which one the following compounds will readily
react with dilute NaOH?
(1) C6H5CH2OH (2) C2H5OH
(3) (CH3)3COH (4) C6H5OH
Ans. (4)
Sol.
OH
+ NaOH
O¯Na+
+ H2O
Stronger ACID than H2O
65. The shape of carbocation is :
(1) trigonal planar (2) diagonal pyramidal
(3) tetrahedral (4) diagonal
Ans. (1)
Sol.
C
H
H
H
Carbocation
Trigonal planar
66. Given below are two statements :
Statement (I) : SN2 reactions are 'stereospecific',
indicating that they result in the formation only one
stereo-isomers as the product.
Statement (II) : SN1 reactions generally result in
formation of product as racemic mixtures. In the
light of the above statements, choose the correct
answer from the options given below :
(1) Statement I is true but Statement II is false
(2) Statement I is false but Statement II is true
(3) Both Statement I and Statement II is true
(4) Both Statement I and Statement II is false
Ans. (3)
Sol. SN2 Inversion
SN1 Racemisation
67. Match List-I with List-II.
List-I
(Reactions)
List-II
(Products)
(A)
NH2
(i) NaNO2 + HCl
(ii) H2O, warm
(I)
OH
CHO
(B)
OH
Na2Cr2O7
H2SO4
(II)
OH
(C)
OH
(i) CHCl3+aq NaOH
(ii) H+
(III)
OH
COOH
(D)
OH
(i) NaOH
(ii) CO2
(iii) H+
(IV)
O
O
Choose the correct answer from the options given
below :
(1) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(2) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
(3) (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
(4) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Ans. (4)
Sol.
NH2
(i) NaNO2 + HCl
(ii) H2O, warm
OH
OH
Na2Cr2O7
H2SO4
O
O
OH
(i) CHCl3+aq NaOH
(ii) H+
OH
CHO
OH
(i) NaOH
(ii) CO2
(iii) H+
OH
COOH
68. Match List-I with List-II.
List-I
(Test)
List-II
(Identification)
(A)
Bayer's test
(I)
Phenol
(B)
Ceric ammonium
nitrate test
(II)
Aldehyde
(C)
Phthalein dye test
(III)
Alcoholic-OH
group
(D)
Schiff's test
(IV)
Unsaturation
Choose the correct answer from the options given
below :
(1) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(2) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(3) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Ans. (4)
Sol. (A) Bayer's test Unsaturation
(B) Ceric ammonium nitrate test Alcoholic-OH group
(C) Phthalein dye test Phenol
(D) Schiff's test Aldehyde
69. Identify the incorrect statements about group 15
elements :
(A) Dinitrogen is a diatomic gas which acts like an
inert gas at room temperature.
(B) The common oxidation states of these
elements are –3, +3 and +5.
(C) Nitrogen has unique ability to form p–p
multiple bonds.
(D) The stability of +5 oxidation states increases
down the group.
(E) Nitrogen shows a maximum covalency of 6.
Choose the correct answer from the options given
below.
(1) (A), (B), (D) only (2) (A), (C), (E) only
(3) (B), (D), (E) only (4) (D) and (E) only
Ans. (4)
Sol. (D) Due to inert pair effect lower oxidation state is
more stable.
(E) Nitrogen belongs to 2nd period and cannot
expand its octet.
70. IUPAC name of following hydrocarbon (X) is :
CH3 – CH – CH2 – CH2 – CH – CH – CH2 – CH3
CH3
CH3
CH3
(X)
(1) 2-Ethyl-3,6-dimethylheptane
(2) 2-Ethyl-2,6-diethylheptane
(3) 2,5,6-Trimethyloctane
(4) 3,4,7-Trimethyloctane
Ans. (3)
Sol.
CH3 – CH – CH2 – CH2 – CH – CH – CH2 – CH3
CH3
CH3
CH3
1
2
3
4
5
6
7
8
2,5,6-Trimethyloctane
71. The equilibrium
22
2 7 4
Cr O 2CrO
is shifted to
the right in :
(1) an acidic medium
(2) a basic medium
(3) a weakly acidic medium
(4) a neutral medium
Ans. (2)
Sol.
OH
22
2 7 4
H
Cr O 2CrO
72. Given below are two statements :
Statement (I) : A Buffer solution is the mixture of
a salt and an acid or a base mixed in any particular
quantities.
Statement (II) : Blood is naturally occurring
buffer solution whose pH is maintained by
2 3 3
H CO / HCO
concentrations.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II is true
(3) Both Statement I and Statement II is false
(4) Statement I is true but Statement II is false
Ans. (1)
Sol. Buffer solution is a mixture of either weak acid /
weak base and its respective conjugate.
Blood is a buffer solution of carbonic acid H2CO3
and bicarbonate
3
HCO
Statement 1 is false but Statement II is true.
73. The correct sequence of acidic strength of the
following aliphatic acids in their decreasing order
is :
CH3CH2COOH, CH3COOH, CH3CH2CH2COOH,
HCOOH
(1) HCOOH > CH3COOH > CH3CH2COOH >
CH3CH2CH2COOH
(2) HCOOH > CH3CH2CH2COOH >
CH3CH2COOH > CH3COOH
(3) CH3CH2CH2COOH > CH3CH2COOH >
CH3COOH > HCOOH
(4) CH3COOH > CH3CH2COOH >
CH3CH2CH2COOH > HCOOH
Ans. (1)
Sol. CH3CH2COOH, CH3COOH, CH3CH2CH2COOH,
HCOOH
The correct order is :
HCOOH > CH3COOH > CH3CH2COOH >
CH3CH2CH2COOH
74. Given below are two statements :
Statement (I) : All the following compounds react
with p-toluenesulfonyl chloride.
C6H5NH2 (C6H5)2NH (C6H5)3N
Statement (II) : Their products in the above
reaction are soluble in aqueous NaOH.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Both Statement I and Statement II is false
(2) Statement I is true but Statement II is false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II is true
Ans. (1)
Sol. Hinsberg test given by 1° amine only.
75. The emf of cell
2
0.001M 0.01M
T1 T1 Cu Cu
is 0.83 V at
298 K. It could be increased by :
(1) increasing concentration of T1+ ions
(2) increasing concentration of both T1+ and Cu2+ ions
(3) decreasing concentration of both T1+ and Cu2+ ions
(4) increasing concentration of Cu2+ ions
Ans. (4)
Sol.
s aq
2
aq s
2
s aq aq s
AnodicReaction T T e 2
CathodicReaction Cu 2e Cu
Overall RedoxReaction 2T Cu 2T Cu
2
o
cell cell 2
T
0.0591
E E log
2Cu
Ecell increases by increasing concentration of [Cu+2]
ions.
76. Identify the correct statements about p-block
elements and their compounds.
(A) Non metals have higher electronegativity than
metals.
(B) Non metals have lower ionisation enthalpy
than metals.
(C) Compounds formed between highly reactive
nonmetals and highly reactive metals are
generally ionic.
(D) The non-metal oxides are generally basic in nature.
(E) The metal oxides are generally acidic or
neutral in nature.
(1) (D) and (E) only (2) (A) and (C) only
(3) (B) and (E) only (4) (B) and (D) only
Ans. (2)
Sol. As electronegativity increases non-metallic nature
increases.
Along the period ionisation energy increases.
High electronegativity difference results in ionic
bond formation.
Oxides of metals are generally basic and that of
non-metals are acidic in nature.
77. Given below are two statements :
Statement (I) : Kjeldahl method is applicable to
estimate nitrogen in pyridine.
Statement (II) : The nitrogen present in pyridine
can easily be converted into ammonium sulphate in
Kjeldahl method.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Both Statement I and Statement II is false
(2) Statement I is false but Statement II is true
(3) Both Statement I and Statement II is true
(4) Statement I is true but Statement II is false
Ans. (1)
Sol. Nitrogen present in pyridine can not be estimated
by Kjeldahl method as the nitrogen present in
pyridine can not be easily converted into
ammonium sulphate.
78. The reaction ;
2(g) (s) (aq) (aq) (s)
1H AgCl H Cl Ag
2
occurs in which of the following galvanic cell :
(1)
2(g) (soln.) (s)
Pt H HCl AgCl Ag
(2)
2(g) (soln.) 3(aq)
Pt H HCl AgNO Ag
(3)
2(g) (soln.) (s)
Pt H KCl AgCl Ag
(4)
(s) (soln.) 3(aq.)
Ag AgCl KCl AgNO Ag
Ans. (3)
Sol. Anodic half cell
Gas – gas ion electrode
2g aq
1H H e
2
Cathodic Reaction
Metal-metal insoluble salt anion electrode
aq s
Ag e Ag
s aq aq
AgCl Ag Cl
s s aq
AgCl e Ag Cl
Overall redox reaction
2g s aq aq s
1H AgCl H Cl Ag
2
Cell Representation
2(g) (sol) (s)
Pt | H | kCl | AgCl | Ag
79. Given below are two statements :
Statement (I) : Fusion of MnO2 with KOH and an
oxidising agent gives dark green K2MnO4.
Statement (II) : Manganate ion on electrolytic
oxidation in alkaline medium gives permanganate
ion.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Both Statement I and Statement II is true
(2) Both Statement I and Statement II is false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Ans. (1)
Sol. MnO2 + 4KOH + O2
fused
2K2MnO4 + 2H2O
Dark green
Electrolytic oxidation in alkaline medium :
At anode :
2
44
MnO MnO e
80. Match List-I with List-II.
List-I
(Complex ion)
List-II
(Spin only magnetic
moment in B.M.)
(A)
[Cr(NH3)6]3+
(I)
4.90
(B)
[NiCl4]2–
(II)
3.87
(C)
[CoF6]3–
(III)
0.0
(D)
[Ni(CN)4]2–
(IV)
2.83
Choose the correct answer from the options given
below :
(1) (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
(2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(3) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(4) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Ans. (3)
Sol. (A) [Cr(NH3)6]3+
Cr3+ : 3d3
n = 3 (unpaired electrons)
3.87 B.M. (II)
(B) [NiCl4]2–
Ni2+ : 3d8
n = 2
2.83 B.M. (IV)
(C) [CoF6]3–
Co3+ : 3d6
n = 4
4.90 B.M. (I)
(D) [Ni(CN)4]2–
Ni2+ : 3d8
n = 0
= 0 B.M. (III)
SECTION-B
81. vap
H
for water is +40.49 kJ mol–1 at 1 bar and
100°C. Change in internal energy for this
vapourisation under same condition is ______ kJ
mol–1. (Integer answer)
(Given R = 8.3 JK–1 mol–1)
Ans. (38)
Sol.
22
H O( ) H O(g)
0
vap
H 40.79 kJ / mole
00
vap vap g
H U n RT
0
vap
8.3 373.15
40.79 U 1000
0
vap
U 40.79 3.0971
37.6929
0
vap
U 38
82. Number of molecules having bond order 2 from
the following molecule is _______.
C2, O2, Be2, Li2, Ne2, N2, He2
Ans. (2)
Sol. C2
(12e–) : 1s2,*1s2,2s2,*2s2
22
xy
2p 2p
B.O. =
84 2
2
O2
(16e–) : 1s2,*1s2,2s2,*2s2,2pz2
22
xy
2p 2p
11
xy
*2p *2p
B.O. =
10 6 2
2
Be2
(8e–) : 1s2,*1s2,2s2,*2s2
B.O. =
44 0
2
Li2
(6e–) : 1s2,*1s2,2s2
B.O. =
421
2
Ne2
(20e–) : 1s2,*1s2,2s2,*2s2,2pz2
22
xy
2p 2p
2 2 2
x y z
*2p *2p *2p
B.O. =
10 10 0
2
N2
(14e–) : 1s2,*1s2,2s2,*2s2
22
xy
2p 2p
2
z
2p
B.O. =
10 4 6
2
He2
(4e–) : 1s2,*1s2
B.O. =
22 0
2
83. Total number of optically active compounds from
the following is _______.
CH3
C
C
CH3
OH
H
OH
H
,
OH
OH
OH
,
CH3 – CH2 – CH2 – CH2 – OH,
CH3 – CH2 – CH – CH3
CH3 – CH2 – CH2 – CH2 – Cl,
Cl
(CH3)2CH – CH2 – CH2 – Cl
Ans. (1)
Sol.
CH3 – CH2 – CH – CH3
Cl
84. The total number of carbon atoms present in
tyrosine, an amino acid, is _______.
Ans. (9)
Sol. Tyrosine
HO
NH2
OH
O
Number of carbon atoms = 9
85. Two moles of benzaldehyde and one mole of
acetone under alkaline conditions using aqueous
NaOH after heating gives x as the major product.
The number of bonds in the product x is
Ans. (9)
Sol.
Ph
H
C = O + CH3–C–CH3 + O = C
Ph
H
O
NaOH/
Ph
H
C = CH–C–CH = C
Ph
H
O
Aldol
condensation
reaction
86. Total number of aromatic compounds among the
following compounds is _______.
, , , , ,
N
Ans. (1)
Sol.
N
87. Molality of an aqueous solution of urea is 4.44 m.
Mole fraction of urea in solution is x × 10–3.
Value of x is _______. (integer answer)
Ans. (74)
Sol. Molality of urea is 4.44 m, that means 4.44 moles
of urea present in 1000 gm of water.
urea
4.44
X1000
4.44 18
= 0.0740
OR
3
74 10
X = 74
88. Total number of unpaired electrons in the complex
ion [Co(NH3)6]3+ and [NiCl4]2– is
Ans. (2)
Sol. Co+3 : 3d6
2,2,2 0,0
2g g
te
Unpaired e– = 0
Ni+2 : 3d8
2,2 2,1,1
2
et
Unpaired e– = 2
89. Wavenumber for a radiation having 5800 Å
wavelength is x × 10 cm–1. The value of x is
_____.
Ans. (1724)
Sol.
(wave no.) =
8
11
5800 10 cm
= 17241
OR
1
1724 10cm
x = 1724
90. A solution is prepared by adding 1 mole ethyl
alcohol in 9 mole water. The mass percent of
solute in the solution is _______ (Integer Answer)
(Given : Molar mass in g mol–1 Ethyl alcohol : 46,
water : 18)
Ans. (22)
Sol. Mass percent of Alcohol
=
Mass of ethyl alcohol 100
Total mass of solution
=
1 46 100
1 46 9 18
=
4600
208
= 22.11 Or 22
